Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 11

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Theorem

Let $p$ be a prime number.

Let $a \in \Z, b \in Z_{> 0}$

Let:

$\forall n \in \N: \exists r_n \in \Z : \dfrac a b - \paren{p^{n + 1} \dfrac {r_n} b} \in \set{0, 1, \ldots, p^{n + 1} - 1}$

Then:

$\forall n \in \N : \dfrac {a - \paren {p^{n + 1} - 1} b } {p^{n + 1} } \le r_n \le \dfrac a {p^{n + 1} }$

Proof

We have:

\(\ds 0\)

\(\le\)

\(\, \ds \dfrac a b - \paren{p^{n + 1} \dfrac {r_n} b} \, \)

\(\, \ds \le \, \)

\(\ds p^{n + 1} - 1\)

\(\ds \leadsto \ \ \)

\(\ds -\paren {p^{n + 1} - 1}\)

\(\le\)

\(\, \ds \paren{p^{n + 1} \dfrac {r_n} b} - \dfrac a b \, \)

\(\, \ds \le \, \)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds -\paren {p^{n + 1} - 1} b\)

\(\le\)

\(\, \ds p^{n + 1} r_n - a \, \)

\(\, \ds \le \, \)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds a - \paren {p^{n + 1} - 1} b\)

\(\le\)

\(\, \ds p^{n + 1} r_n \, \)

\(\, \ds \le \, \)

\(\ds a\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {a - \paren {p^{n + 1} - 1} b} {p^{n + 1} }\)

\(\le\)

\(\, \ds r_n \, \)

\(\, \ds \le \, \)

\(\ds \dfrac a {p^{n + 1} }\)

The result follows.

$\blacksquare$